Question

In a factory, 2000 kg of metal needs to be lifted by an engine through a distance of 12 m in 1 minute. Find the minimum horsepower of the engine to be used.

Answer 1

Minimum power is necessary. That's 5.25 horsepower.

Explanation:

Step 1:

Given values in the question  

The mass  of metal m = 2000 kg.

Lift height   h = 12 m.

Time taken t = 1 min = 60 sec.

So, here is the system's limited function as metal raised gradually.

Step 2:

Now work done against the gravity ,

= mgh

= 2000 × 9.8 × 12

= 19600 × 12

W= 235200 J.

Step 3:

Now for the Power, P = (work) / (time)

P = (w) / (t)

P = (235200) / 60

P = 3920 W.

Changing in hp.

P =3920/746  [1 hp = 746 W.]

P = 5.25 hp

Minimum power is therefore necessary. That's 5.25 horsepower.

Answer 2

Required Solution :-

Here we've been given with the mass of metal , distance covered (displacement) by metal , and the time taken by it.

Here only gravitational force would be acting on the metal towards the centre of earth. So let's calculate the gravitational force !

As we know :

• F = mg

We have :

• m = 2000 kg
• g = 10 (Approx)

→ F = 2000 × 10

→ F = 20000 N

Calculating the Work done :

• W = FS cos theta

Here,

• F is force applied on metal
• S is displacement occurred
• Theta is the angle between force and displacement. (Theta would be 0° as there's no angle between them)

→ W = 20000 × 12 × cos0°

→ W = 20000 × 12 × 1

→ W = 20000 × 12

→ W = 240000 J

At last we would be finding out the power consumed inorder to do so. Then we would convert that power into horsepower.

As we know that :

• Power (P) = Work done (W) / Time (T)

We have :

• W = 240000 J
• T = 60s

Substituting the values :

→ P = 240000 / 60

→ P = 24000 / 6

→ P = 12000 / 3

→ P = 4000 Watts

Converting unit :

• As we know that 1H.P. = 746 W

→ P = 4000 / 746

→ P = 5.3 H.P.